I suggest that instead of using f = str2func(['@(x)' vectorize(f)]); that you use f = matlabFunction(f); However, you could just comment out all of your handling of f, as you never use f after you make it into a function handle. You have z(t)=integral(y,-t,t); At this point your y is a symbolic variable. integral() cannot be applied to symbolic variables: you would need to use int(y, -t, t) -- which is a value you can easily predict will be 0, since the integral of y with respect to y over y = a to y = b is 1/2 b^2 - 1/2 a^2 and with a = -t and b = -t that is going to be 1/2 t^2 - 1/2 (-t)^2 which is going to be 0.

Use the buffer function to segment the signal into 1-second increments. Since you want 1-second samples, the vector length is simply ‘Fs’, so — y = randn(73113,1); % Create Signal To Test Code (One Channel) fs = 8192; % Sampling Frequency (Also Frame Length) N = size(y,1); % Signal Length time = linspace(0, N, N-1)/fs; % Time Vector y_b = bandpass(y,[300 2000],fs); % Filter Signal y_seg = buffer(y_b,fs); % Matrix Of Signal Segments y_pwr = sum(y_seg.^2); % Calculate Power figure for k = 1:numel(y_pwr)-1 tv = time((1:fs)+fs*(k-1)); subplot(5,2,k) plot(tv,y_seg(:,k)) grid title(sprintf('Segment %d Power %.1f',k,y_pwr(k))) end subplot(5,2,k+1) tv = time(fs*k:end); plot(tv,y_seg(1:numel(tv),k+1)) grid title(sprintf('Segment %d Power %.1f',k+1,y_pwr(k+1))) EDIT — (28 Apr 2021 at 12:40) Corrected typoographical error.

The Matlab function int needs a symbolic expression as an integrand, and a symbolic variable as an integration variable. In your example s and t are both vectors, and you can't use int in this case. If you want to compute the power or energy of discrete time signals, then you need to use the corresponding definitions:

$$E_x=\sum_{n=-\infty}^{\infty}|x_n|^2\\ P_x=\lim_{N\rightarrow\infty}\frac{1}{2N+1}\sum_{n=-N}^N|x_n|^2$$

In practice you will need to consider finite length signals unless you have an analytic formula for $x_n$ and some software package that can deal with the analytic evaluation of infinite sums. Obviously, finite length signals always have finite energy and $P_x=0$ according to above definition.